Application of Maximize: The cuboid constrained to the ellipsoid in $\mathbb{R}^3$

Question

Since I am a student of Mathematics I enjoy to apply MMA to problems that I have a solid understanding in. The following would be such a problem:

Maximize $f: \mathbb{R}^3 \to \mathbb{R}$ given by $f(x,y,z):= 8xyz$ constrained to the ellipsoid $$ E:= \left\lbrace (x,y,z) \in \mathbb{R}^3: \frac{x^2}{a^2}+ \frac{y^2}{b^2}+ \frac{z^2}{c^2}=1 \right\rbrace $$ where $a,b,c>0$

Under the realistic assumptions that $x,y,z >0$ are all positive (because else wise the cuboid vanishes) this is an easy application by hand on paper with a little help from multivariate Calculus, namely the Lagrange multipliers.

After sume juggling with the gradients one obtains the solutions of the maximum $$x_0= \frac{a}{\sqrt{3}}, \ y_0 = \frac{b}{\sqrt{3}}, \ z_0 = \frac{c}{\sqrt{3}} \implies V_\max=f(x_0,y_0,z_0)= \frac{8abc}{3\sqrt{3}} $$

---

In Mathematica however I struggle to implement this problem:

f[x_,y_,z_]:= 8 x y z (* a naive R^3 -> R function *)

Then indeed I use the Maximize function with the appropriate constraints:

Maximize[{f[x,y,z], (x/a)^2+(y/b)^2+(z/c)^2 ==1 && a> 0 && b >0 && c>0}, {x,y,z}]

MMA manages to find the same Maximal volume as I did by hand, but for the variables $x,y,z$ it finds very peculiar results:

While the volume is correct, I wonder how I get the more realistic $x,y,z$ results as I did by hand.

Addendum: More realistically I would have rather defined a function as

g[x_ /; x>0, y_ /; y>0, z_ /; z>0]:= 8 x y z

But in this case Maximize won't even execute

Answer 1

Simplifying complicated results sometimes takes a long time, so many Mathematica functions do not do so by default. Or they do it under a time constraint. Such is the case here, and in addition, FullSimplify is needed to simplify the nested Root objects. Note we need to add the constraints as assumptions to get the desired simplification.

res = Maximize[{f[x,y,z], (x/a)^2 + (y/b)^2 + (z/c)^2 == 1 &&
        a > 0 && b > 0 && c > 0}, {x, y, z}]
FullSimplify[res, a > 0 && b > 0 && c > 0]
(*
  {(a b c)/(3 Sqrt[3]), {x -> -(a/Sqrt[3]), y -> -(b/Sqrt[3]), z -> c/Sqrt[3]}}
*)

P.S. There are four equivalent answers. If you want x, y, and x to be all positive, add more constraints. (Mathematica probably found the "first" one in the code above, and since negatives are sorted before positives, we got the solution that starts with two minuses.)

res = Maximize[{f[x y z], (x/a)^2 + (y/b)^2 + (z/c)^2 == 1 &&
        a > 0 && b > 0 && c > 0 && x > 0 && y > 0 && z > 0}, {x, y, z}];